Question:
<p>Let the product of the focal distances of the point <span class="math-tex">\(\left(\sqrt{3}, \frac{1}{2}\right)\)</span> on the ellipse <span class="math-tex">\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,(a \gt b)\)</span>, be <span class="math-tex">\(\frac{7}{4}\)</span>. Then the absolute difference of the eccentricities of two such ellipses is</p>
<p style="display:inline"><span class="math-tex">\(\frac{1-\sqrt{3}}{\sqrt{2}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{3-2 \sqrt{2}}{3 \sqrt{2}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1-2 \sqrt{2}}{\sqrt{3}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{3-2 \sqrt{2}}{2 \sqrt{3}}\)</span></p>
Step-by-Step Solution
Key Concept: Utilize the product of focal distances formula $(a^2 - e^2x_1^2)$ alongside the ellipse equation and the eccentricity relation $b^2 = a^2(1-e^2)$ to solve for $e^2$.
<p>Let Product of focal distances <span class="math-tex">$=(a+ \left.e x_{1}\right)\left(a-e x_{1}\right)$</span><br />
<span class="math-tex">$=a^{2}-e^{2} x_{1}^{2}=a^{2}-e^{2}$</span><br />
<span class="math-tex">$=a^{2}-3 e^{2}=\frac{7}{4} \Rightarrow a^{2}=\frac{7}{4}+3 e^{2}$</span><br />
<span class="math-tex">$\Rightarrow 4 a^{2}=7+12 e^{2}$</span><br />
<span class="math-tex">$\left(\sqrt{3}, \frac{1}{2}\right)$</span> lies on <span class="math-tex">$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$</span><br />
<span class="math-tex">$\therefore \frac{3}{a^{2}}+\frac{1}{4 b^{2}}=1$</span><br />
<span class="math-tex">$\Rightarrow \frac{3}{a^{2}}+\frac{1}{4\left(a^{2}\right)\left(1-e^{2}\right)}=1 \left[\because b^{2}=a^{2}\left(1-e^{2}\right)\right]$</span><br />
<span class="math-tex">$\Rightarrow 12\left(1-e^{2}\right)+1=4 a^{2}\left(1-e^{2}\right)$</span><br />
<span class="math-tex">$\Rightarrow 13-12 e^{2}=\left(7+12 e^{2}\right)\left(1-e^{2}\right)$</span><br />
<span class="math-tex">$\Rightarrow 13-12 e^{2}=7-7 e^{2}+12 e^{2}-12 e^{4}$</span><br />
<span class="math-tex">$\Rightarrow 12 e^{4}-17 e^{2}+6=0$</span><br />
<span class="math-tex">$\therefore e^{2}=\frac{17 \pm \sqrt{289-288}}{24}=\frac{17 \pm 1}{24}$</span><br />
<span class="math-tex">$=\frac{3}{4} \& \frac{2}{3}$</span><br />
<span class="math-tex">$\therefore e=\frac{\sqrt{3}}{2} \& \sqrt{\frac{2}{3}}$</span><br />
∴ difference <span class="math-tex">$=\frac{\sqrt{3}}{2}-\sqrt{\frac{2}{3}}=\frac{3-2 \sqrt{2}}{2 \sqrt{3}}$</span></p>
Correct Answer: D